Respuesta :

  • The total resistance in the circuit is 16 Ohms.
  • The total current in the circuit is 0.5 Ampere.
  • The current at [tex]R_1[/tex] is 0.5 Ampere.
  • The current at [tex]R_3[/tex] is 0.5 Ampere.
  • The voltage drop at[tex]R_1[/tex] is 4 volts.
  • The voltage drop at [tex]R_2[/tex] is 2.5  volts.
  • The voltage drop at [tex]R_3[/tex] is 1.5 volts.
  • The total power consumed by the circuit is 4watts
  • The power consumed at [tex]R_1[/tex] is 2 watts
  • The power consumed at [tex]R_2[/tex] is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

[tex]R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega[/tex]

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

  • The total resistance in the circuit  = [tex]R_{eq}[/tex]

[tex]R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega[/tex]

  • The total current in the circuit = I

[tex]V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A[/tex] (Ohm's law)

  • For series combinations, the current in each resistor remains the same.

So, the current in [tex]R_1, R_2 \&R_3[/tex]:

[tex]I_1= I_2= I_3=I=0.5 A\\[/tex]

  • The voltage drop across at [tex]R_1[/tex] = [tex]V_1[/tex]

The current across  [tex]R_1[/tex] = I = 0.5 A

[tex]V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V[/tex]

  • The voltage drop across at [tex]R_2[/tex] =[tex]V_2[/tex]

The current across  [tex]R_2[/tex] = I = 0.5 A

[tex]V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V[/tex]

  • The voltage drop across at [tex]R_3[/tex] = [tex]V_3[/tex]

The current across  [tex]R_3[/tex] = I = 0.5 A

[tex]V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V[/tex]

  • The total power consumed by circuit:

[tex]P= V\times I \\\\= 0.5 A\times 8 V = 4 watt[/tex]

  • Power consumed at [tex]R_1[/tex]:

[tex]P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt[/tex]

  • Power consumed at [tex]R_2[/tex]:

[tex]P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt[/tex]

  • Power consumed at [tex]R_3[/tex]:

[tex]P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt[/tex]

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